7. (30 points) Suppose a worker needs to process 36 orders. The time to process

Question

7. (30 points) Suppose a worker needs to process 36 orders. The time to process each item has a Pareto- Levy distribution with a mean of 10 minutes and a variance of 100 minutes2. Assume that the processing times of the jobs are i.i.d. (a) Approximately what is the probability that the worker finishes in less than 7 hours? (b) What is the probability that the worker finishes more than 4 hours? (c) Roughly how much time would the worker need so that the probability of finishing within that time be 97.5%?

Explanation / Answer

ahere expected time to finish all orders =10*36 =360 minutes

and std deviaiton =(100*36)1/2 =60

a) probability that worker finishes in less than 7 hours =P(X<420) =P(Z<(420-360)/60)=P(Z<1) =0.8413

b) P(X>240)=P(Z>(240-360)/60)=P(Z>-2)=0.97725

c) for 96.5 percentile ; z score =1.96

hence corresponding time =360+1.96*60=477.60

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